A Nice Factorial Equation | n!=2^n

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[yoav]

Just check n=0, 1, 2 if n>=3 there is the prime 3 in n! 😀

[wtspman]

The real value solution at ~3.4 represents the crossing point where n! switches from being less than 2^n to being greater than 2^n.

[bobbyheffley]

By the intermediate value theorem, x!=2^x somewhere between 3 and 4.

[yoav]

The solution n=3.459 can be found numerically. For example by neuton method. We have gamma(x+1)=2^x so f(x)= gamma(x+1)-2^x. Now we use x(n+1)=x(n)- f(x(n))/f'(x(n)) until covergence.

[ronbelanger]

Mathematical induction is built into the very core of the natural numbers. Your proof was very rigorous. Thanks for your work!

[lawrencejelsma]

Digital sampling mathematics in electrical engineering would have of course found the n=0 solution (without n<0) but it would have also found the later crossover minumum ansolute value difference in a numerical comparitor. Using some sort of + voltage and - voltage comparitor circuitry it would have approximated (knowing error but using integers as sampling of data) n = 3 as another solution where 2^3 - 3! = 2 and then the reversal voltage 2^4 - 4! = -8. The minimal voltage change in the comparitor circuit is then the one of smallest voltage difference. Usually millivolts or microvolts comparitor circuits continually check two input diffferences in voltages to signal to a controller switch to get ready and switch on or off some circuitry.

[luisoswaldoramirezzevallos]

You did, I remmenber my time' university. Thanks.

[TypoKnig]

I was going to suggest using Stirling’s approximation for the factorial function as a way to start determining the x value between 3 and 4, but I remembered that Stirling’s approximation is not great for n < 10. The way to go would be to use the Gamma function, using iterations. Is there a useful form of an inverse Gamma function?

[jpolowin]

Not to be confused with "n != 2^n" with the computer-language "!=" meaning '≠'. That is true for all real values of n; I don't have the working math brain to check complex values.

[alextang]

For n>=3, left hand side has 3 as a factor but right has not. Thus, n>=3, there is no root. 😎😎😎😎😎😎

[goldfing]

I only find n = 0, because 0! = 1 = 2^0. I tested the following integers:
0! = 1 and 2^0 = 1
1! = 1 and 2^1 = 2
2! = 2 and 2^2 = 4
3! = 6 and 2^3 = 8
4! = 24 and 2^4 = 16
5! = 120 and 2^5 = 32
For n >= 4, n! is greater than 2^n, because for larger numbers, the faculty grows more quickly, because when increasing n by one, the additional factor in the product is 2 for 2^n but greater than 2 for n! You cpild probably prove this by induction.

[seanfraser]

n=0 is the only solution.

You can verify that n=1 and n=2 are not solutions, and 3 divides n! for n>=3 and thus it can never be a power of 2 again.

[StuartSimon]

The gamma function has asymptotes at the negative integers.

[maxvangulik]

Int{0, infinity}(x^ne^-xdx)*2^-n
=Int{0, infinity}((x/2)^ne^-xdx)=1
Would this provide any insight?

[JanThomas-gy]

Not a mathematician, just a lowly mechanical engineer (retired) and I must say that factorials never came up once at work, but surely 0! = 0. Just checked a few utube vids on the subject, and just doesn't seem right.
Also, why is the factorial graph continuous?
Anyway, really enjoy these fun SyberMath quiz questions. Please don't take my queries as a white glove to the face of mathematics. Cheers.

[mcwulf]

Why induction? 2^n is never divisible by 3, so n < 3.

[miguelcerna]

Take the logarithm base 2 of both sides:

log2(n!) = n * log2(2)

Since log2(2) = 1, the equation becomes:

log2(n!) = n

Use Stirling's approximation for the factorial:

n! ≈ √(2πn) * (n / e)^n

Substituting this into the equation:

log2(√(2πn) * (n / e)^n) = n

Simplify further:

0.5 * log2(2πn) + n * log2(n / e) = n

0.5 * log2(2πn) = 0.5 * n * log2(n / e)

log2(2πn) = n * log2(n / e)

Solve for n:

log2(2πn) = n * (log2(n) - log2(e))

Divide both sides by n:

log2(2π) + log2(n) = log2(n) - log2(e)

Move the log2(n) terms to one side:

log2(2π) + log2(e) = 0

This means that there is no solution that satisfies the original equation n! = 2^n. So, there is no integer value of n that simultaneously satisfies this equation.

[anestismoutafidis]

Inf! = 2^inf <=> n1=inf; \ 0!=2^0 <=> n2=0

[warever]

It would be more interesting if task would really involve gamma-function for solution.
Current task is boring, sorry.

[geoffreyparfitt]

I'm sorry but treating x! as a continuous function seems rediculess to me.